We already know that the solution is (9, -3/5), let's show that
[tex]\frac{1}{10}x+\frac{1}{3}y=\frac{7}{10}[/tex]Let's plug our solution into the equation
[tex]\begin{gathered} \frac{1}{10}\cdot9+\frac{1}{3}\cdot(-\frac{3}{5})=\frac{7}{10} \\ \\ \frac{9}{10}-\frac{3}{15}=\frac{7}{10} \\ \\ \frac{9\cdot15-3\cdot10}{10\cdot15}=\frac{7}{10} \\ \\ \frac{135-30}{150}=\frac{7}{10} \\ \\ \frac{105}{150}=\frac{7}{10} \\ \\ \frac{21}{30}=\frac{7}{10} \\ \\ \boxed{\dfrac{7}{10} =\dfrac{7}{10} } \end{gathered}[/tex]The second equation is true!
